Advertisements
Advertisements
प्रश्न
If `(y + z)/a = (z + x )/b = (x + y)/c` then show that `x/[b + c - a ] = y/[c + a - b] = z/(a + b - c)`
उत्तर
`(y + z)/a = (z + x)/b = (x + y)/ c`
By invertendo,
`a/(y + z)= b /(z + x )= c/(x + y)`
`a/(y + z)= b /(z + x)= c/(x + y) = (b + c - a)/(z + x + x + y - y - z)` ...( Theorem of equal ratios)
⇒ `a/(y + z)= b /(z + x)= c/(x + y) = (b + c - a)/[2x]` ...(1)
Now,
`a/(y + z)= b /(z + x )= c/(x + y) = [c + a - b]/[x + y + y + z - z -x]` ...( Theorem of equal ratios)
⇒ `a/(y + z)= b /(z + x)= c/(x + y) = (c + b - a)/[2y]` ...(2)
Also,
`a/(y + z)= b /(z + x)= c/(x + y) = [a + b - c]/[y + z + z + x - x - y ]` ...( Theorem of equal ratios)
⇒ `a/( y + z)= b /(z + x)= c/(x + y) = (a + b - c)/[2z]` ...(3)
From (1), (2) and (3), we have
`[b + c - a]/(2x) = [c + a - b]/( 2y) = (a + b - c)/(2z)`
⇒ `[b + c - a]/x = [c + a - b]/y = (a + b - c)/z`
by invertendo,
⇒ `x/[b + c - a]= y/[c + a - b]= z/(a + b - c)`
APPEARS IN
संबंधित प्रश्न
Fill in the blanks of the following.
`x/7 = y/3 = (3x + 5y)/("_____") = (7x -9y)/("_____")`
If a(y + z) = b(z + x) = c(x + y) and out of a, b, c no two of them are equal then show that,
`(y - z)/[a ( b - c )] = ( z - x)/[ b ( c - a)] = ( x - y)/[c ( a - b )]`
If `x/[3x - y -z] = y/[3y - z -x] = z/[3z -x -y]` and x + y + z ≠ 0 then show that the value of each ratio is equal to 1.
If `[ax + by]/( x + y) = ( bx + az )/(x + z) = (ay + bz)/[y + z]` and `x + y + z ≠ 0 ` then show that `[a + b]/2`.
If `{ 3x - 5y }/ ( 5z + 3y ) = ( x + 5z )/( y - 5x ) = ( y - z )/ ( x - z )`then show that every ratio = `x/y`.
Solve.
`[ 16x^2 - 20x +9]/[ 8x^2 + 12x + 21] = ( 4x - 5 )/( 2x + 3)`
Solve.
`( 5y^2 + 40y - 12)/( 5y + 10y^2 - 4) = ( y + 8)/( 1 + 2y)`
Solve:
`[12x^2 + 18x + 42]/[ 18 x^2 + 12x +58 ] = [ 2x + 3]/[ 3x + 2]`
If `[ 2x - 3y ]/[ 3z + y] = [ z - y ]/[ z - x ] = [ x + 3z ]/[ 2y - 3x]` then prove that every ratio = `x/y`.
If `[by + cz ]/[b^2 + c^2] = [cz + ax]/[c^2 + a^2] = [ax + by]/[a^2 + b^2]` then prove that `x/a= y/b = z/c`
